# How do you solve this integral?

• Aug 8th 2008, 05:40 PM
mojo0716
How do you solve this integral?
$\int_{1}^{2} 1/(x+1)^2 \: dx$

not sure how to solve an integral with a quantity squared in the denominator. I'm new to integration. do you foil the bottom or rewrite it as (x+1)^-2? or what?

thanks
• Aug 8th 2008, 05:44 PM
Chris L T521
Quote:

Originally Posted by mojo0716
$\int_{1}^{2} 1/(x+1)^2 \: dx$

Make the u substitution $u=x+1\implies \,du=\,dx$

Thus, the integral transforms into $\int_{1}^{2} \frac{\,dx}{(x+1)^2}\implies\int_2^3\frac{\,du}{u^ 2}$

I'm sure you can take it from here.

--Chris
• Aug 8th 2008, 05:46 PM
mojo0716
Quote:

Originally Posted by Chris L T521
Make the u substitution $u=x+1\implies \,du=\,dx$

Thus, the integral transforms into $\int_{1}^{2} \frac{\,dx}{(x+1)^2}\implies\int_2^3\frac{\,du}{u^ 2}$

I'm sure you can take it from here.

--Chris

ah thanks
• Aug 8th 2008, 05:52 PM
mojo0716
The solution manual has it like this, though, and I wonder what they did? Looks like they didn't use any substitution.

$\int_{1}^{2} 1/(x+1)^2 \: dx$ = (-1/(x+1)) from 1 to 2

after that I know what to do, but I don't know how they got = (-1/(x+1)) from 1 to 2

thanks
• Aug 8th 2008, 06:01 PM
Chris L T521
Quote:

Originally Posted by mojo0716
The solution manual has it like this, though, and I wonder what they did?

$\int_{1}^{2} 1/(x+1)^2 \: dx$ = (-1/x+1) from 1 to 2

thanks

There way is correct, and so is mine.

When we integrate $\frac{1}{(x+1)^2}$, we still need to apply a substitution.

The integral would become $\int\frac{\,du}{u^2}=-\frac{1}{u}$

What they did was re-substitute u back into the antiderivative:

$-\frac{1}{u}=-\frac{1}{x+1}$, and then they evaluated from 1 to 2.

However, the way I did it, I eliminated the need to re-substitute u back into the equation.

I did this when I converted the integration limits.

Since $u=x+1$ and the x limits of integration were from 1 to 2, I saw that the u limits would be from $u=1+1=\color{red}\bold 2$ to $u=2+1=\color{red}\bold 3$

So when you evaluate, we see that

$\int_1^2\frac{\,dx}{(x+1)^2}=\int_2^3\frac{\,du}{u ^2}=\color{red}\boxed{\frac{1}{6}}$

I hope this clarified things! If you have questions, feel free to ask. We're here to help you!

--Chris
• Aug 8th 2008, 06:09 PM
mojo0716
Quote:

Originally Posted by Chris L T521
There way is correct, and so is mine.

When we integrate $\frac{1}{(x+1)^2}$, we still need to apply a substitution.

The integral would become $\int\frac{\,du}{u^2}=-\frac{1}{u}$

What they did was re-substitute u back into the antiderivative:

$-\frac{1}{u}=-\frac{1}{x+1}$, and then they evaluated from 1 to 2.

However, the way I did it, I eliminated the need to re-substitute u back into the equation.

I did this when I converted the integration limits.

Since $u=x+1$ and the x limits of integration were from 1 to 2, I saw that the u limits would be from $u=1+1=\color{red}\bold 2$ to $u=2+1=\color{red}\bold 3$

So when you evaluate, we see that

$\int_1^2\frac{\,dx}{(x+1)^2}=\int_2^3\frac{\,du}{u ^2}=\color{red}\boxed{\frac{1}{6}}$

I hope this clarified things! If you have questions, feel free to ask. We're here to help you!

--Chris

I see.. thanks
• Aug 8th 2008, 06:13 PM
ticbol
Quote:

Originally Posted by mojo0716
The solution manual has it like this, though, and I wonder what they did? Looks like they didn't use any substitution.

$\int_{1}^{2} 1/(x+1)^2 \: dx$ = (-1/(x+1)) from 1 to 2

after that I know what to do, but I don't know how they got = (-1/(x+1)) from 1 to 2

thanks

That is because they did not use substitution. They integarated it directly.

INT[1 / (1+x)^2]dx
= INT[(1 +x)^(-2)]dx
= 1/(-1) *(1 +x)^(-1) +C
= -(1 +x)^(-1) +C
= -1 / (1 +x) +C

And so the boundaries for the integration for dx remained from x=1 to x=2.